If $\sqrt{1-\cos^2x}-\sqrt{1+\sin^2x}=k$, find $\sqrt{1-\cos^2x}+\sqrt{1+\sin^2x}.$ Additional data added

Question

If $\sqrt{1-\cos^2x}-\sqrt{1+\sin^2x}=k$, find $\sqrt{1-\cos^2x}+\sqrt{1+\sin^2x}$

I raised $\sqrt{1-\cos^2x}+\sqrt{1+\sin^2x}$ to the second power in order to express $k$:

\begin{align}&\color{white}=1-\cos^2x+2\sqrt{1-\cos^2x}\sqrt{1+\sin^2x}+1+\sin^2x\\&=1-\cos^2x+\sqrt{1+\sin^2x}(2\sqrt{1-\cos^2x}-\sqrt{1+\sin^2x})\\&=1-\cos^2x+\sqrt{1+\sin^2x}2k.\end{align}

Since it doesn't give the answer, I also raised the second expression equalling $k$ to the second power. That didn't work out. The variants are:

A)$1.5k$ B)$2k$ C)$\frac{2}{k}$ D)$-k$ E)$-\frac{1}{k}$

I have solved it in the way Fred did (having learnt from him) but as is obvious there is NOT such an answer in the variants above.

The correct answer is E)$-\frac{1}{k}$ (according to the answer tables in the book).

How to get to that correct answer E)???

Answer 1

It's obvious that $k<0$.

Let $\sqrt{1-\cos^2x}+\sqrt{1+\sin^2x}=p$.

Thus, $\sqrt{1+\sin^2x}=\frac{p-k}{2}$ and $\sqrt{1-\cos^2x}=\frac{p-k}{2}$.

Id est, $$\left(\frac{p-k}{2}\right)^2-\left(\frac{p+k}{2}\right)^2=1+\sin^2x-1+\cos^2x$$ or $$-pk=1$$ or $$p=-\frac{1}{k}.$$ Done!

Answer 2

$k(\sqrt{1-\cos^2x}+\sqrt{1+\sin^2x})=(\sqrt{1-\cos^2x}-\sqrt{1+\sin^2x})(\sqrt{1-\cos^2x}+\sqrt{1+\sin^2x})=1-\cos^2x-(1+\sin^2x)=-1$.

Answer 3

Let the answer be x.Then get the question and answer in the form (a+b)(a-b) .As you know formula for this.In solving we get -1/k.